\(\text{Đ}\text{ặt}:C_xH_yO_z\left(x,y,z:nguy\text{ê}n,d\text{ươ}ng\right)\\ B1:n_{H_2O}=0,05\left(mol\right)\\ B2:n_{CO_2}=0,05\left(mol\right)\\ x=\dfrac{n_{CO_2}}{n_A}=\dfrac{0,05}{0,025}=2;y=\dfrac{2.n_{H_2O}}{n_A}=\dfrac{2.0,05}{0,025}=4\\ BT.Oxi:n_{O\left(trong.A\right)}+2n_{O_2}=2n_{CO_2}+n_{H_2O}\\ \Leftrightarrow n_{O\left(trongA\right)}+2.0,05=2.0,05+0,05\\ \Leftrightarrow n_{O\left(trongA\right)}=0,05\left(mol\right)\\ \rightarrow z=\dfrac{n_{O\left(trongA\right)}}{n_A}=\dfrac{0,05}{0,025}=2\\ \Rightarrow CTPT.A:C_2H_4O_2\)
\(Đ ặt : C x H y O z ( x , y , z : n g u y ê n , d ươ n g ) B 1 : n H 2 O = 0 , 05 ( m o l ) B 2 : n C O 2 = 0 , 05 ( m o l ) x = n C O 2 n A = 0 , 05 0 , 025 = 2 ; y = 2. n H 2 O n A = 2.0 , 05 0 , 025 = 4 B T . O x i : n O ( t r o n g . A ) + 2 n O 2 = 2 n C O 2 + n H 2 O ⇔ n O ( t r o n g A ) + 2.0 , 05 = 2.0 , 05 + 0 , 05 ⇔ n O ( t r o n g A ) = 0 , 05 ( m o l ) → z = n O ( t r o n g A ) n A = 0 , 05 0 , 025 = 2 ⇒ C T P T . A : C 2 H 4 O 2\)
